# Minimum steps to reach any of the boundary edges of a matrix | Set 1

• Difficulty Level : Medium
• Last Updated : 02 Jun, 2021

Given an N X M matrix, where ai, j = 1 denotes the cell is not empty, ai, j = 0 denotes the cell is empty and ai, j = 2, denotes that you are standing at that cell. You can move vertically up or down and horizontally left or right to any cell which is empty. The task is to find the minimum number of steps to reach any boundary edge of the matrix. Print -1 if not possible to reach any of the boundary edges.
Note: There will be only one cell with value 2 in the entire matrix.
Examples:

```Input: matrix[] = {1, 1, 1, 0, 1}
{1, 0, 2, 0, 1}
{0, 0, 1, 0, 1}
{1, 0, 1, 1, 0}
Output: 2
Move to the right and then move
upwards to reach the nearest boundary
edge.

Input: matrix[] = {1, 1, 1, 1, 1}
{1, 0, 2, 0, 1}
{1, 0, 1, 0, 1}
{1, 1, 1, 1, 1}
Output: -1 ```

Approach: The problem can be solved using a Dynamic Programming approach. Given below is the algorithm to solve the above problem.

• Find the position which has ‘2’ in the matrix.
• Initialize two 2-D arrays of size same as the matrix. The dp[][] which stores the minimum number of steps to reach any index i, j and vis[][] marks if any particular i, j position has been visited or not previously.
• Call the recursive function which has the base case as follows:
1. if the traversal at any point reaches any of the boundary edges return 0.
2. if the points position n, m has stored the minimum number of steps previously, then return dp[n][m].
• Call the recursion again with all possible four moves that can be done from the position n, m. The moves are only possible if mat[n][m] is 0 and the position has not been visited previously.
• Store the minimal of the four moves.
• If the recursion returns any value less than 1e9, which we had stored as the maximum value, then there is an answer, else it does not have any answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to find Minimum steps` `// to reach any of the boundary` `// edges of a matrix` `#include ` `using` `namespace` `std;` `#define r 4` `#define col 5`   `// Function to find out minimum steps` `int` `findMinSteps(``int` `mat[r][col], ``int` `n, ``int` `m, ``int` `dp[r][col], ``bool` `vis[r][col])` `{` `    ``// boundary edges reached` `    ``if` `(n == 0 || m == 0 || n == (r - 1) || m == (col - 1)) {` `        ``return` `0;` `    ``}`   `    ``// already had a route through this` `    ``// point, hence no need to re-visit` `    ``if` `(dp[n][m] != -1)` `        ``return` `dp[n][m];`   `    ``// visiting a position` `    ``vis[n][m] = ``true``;`   `    ``int` `ans1, ans2, ans3, ans4;`   `    ``ans1 = ans2 = ans3 = ans4 = 1e9;`   `    ``// vertically up` `    ``if` `(mat[n - 1][m] == 0) {` `        ``if` `(!vis[n - 1][m])` `            ``ans1 = 1 + findMinSteps(mat, n - 1, m, dp, vis);` `    ``}`   `    ``// horizontally right` `    ``if` `(mat[n][m + 1] == 0) {` `        ``if` `(!vis[n][m + 1])` `            ``ans2 = 1 + findMinSteps(mat, n, m + 1, dp, vis);` `    ``}`   `    ``// horizontally left` `    ``if` `(mat[n][m - 1] == 0) {` `        ``if` `(!vis[n][m - 1])` `            ``ans3 = 1 + findMinSteps(mat, n, m - 1, dp, vis);` `    ``}`   `    ``// vertically down` `    ``if` `(mat[n + 1][m] == 0) {` `        ``if` `(!vis[n + 1][m])` `            ``ans4 = 1 + findMinSteps(mat, n + 1, m, dp, vis);` `    ``}`   `    ``// minimum of every path` `    ``dp[n][m] = min(ans1, min(ans2, min(ans3, ans4)));` `    ``return` `dp[n][m];` `}`   `// Function that returns the minimum steps` `int` `minimumSteps(``int` `mat[r][col], ``int` `n, ``int` `m)` `{` `    ``// index to store the location at` `    ``// which you are standing` `    ``int` `twox = -1;` `    ``int` `twoy = -1;`   `    ``// find '2' in the matrix` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``for` `(``int` `j = 0; j < m; j++) {` `            ``if` `(mat[i][j] == 2) {` `                ``twox = i;` `                ``twoy = j;` `                ``break``;` `            ``}` `        ``}` `        ``if` `(twox != -1)` `            ``break``;` `    ``}`   `    ``// Initialize dp matrix with -1` `    ``int` `dp[r][col];` `    ``memset``(dp, -1, ``sizeof` `dp);`   `    ``// Initialize vis matrix with false` `    ``bool` `vis[r][col];` `    ``memset``(vis, ``false``, ``sizeof` `vis);`   `    ``// Call function to find out minimum steps` `    ``// using memoization and recursion` `    ``int` `res = findMinSteps(mat, twox, twoy, dp, vis);`   `    ``// if not possible` `    ``if` `(res >= 1e9)` `        ``return` `-1;` `    ``else` `        ``return` `res;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `mat[r][col] = { { 1, 1, 1, 0, 1 },` `                      ``{ 1, 0, 2, 0, 1 },` `                      ``{ 0, 0, 1, 0, 1 },` `                      ``{ 1, 0, 1, 1, 0 } };`   `    ``cout << minimumSteps(mat, r, col);` `}`

## Java

 `// Java program to find Minimum steps` `// to reach any of the boundary` `// edges of a matrix` `class` `Solution` `{` `static` `final` `int`  `r=``4``,c=``5``;`   `// Function to find out minimum steps` `static` `int` `findMinSteps(``int` `mat[][], ``int` `n, ``int` `m, ``int` `dp[][], ``boolean` `vis[][])` `{` `    ``// boundary edges reached` `    ``if` `(n == ``0` `|| m == ``0` `|| n == (r - ``1``) || m == (c - ``1``)) {` `        ``return` `0``;` `    ``}` ` `  `    ``// already had a route through this` `    ``// point, hence no need to re-visit` `    ``if` `(dp[n][m] != -``1``)` `        ``return` `dp[n][m];` ` `  `    ``// visiting a position` `    ``vis[n][m] = ``true``;` ` `  `    ``int` `ans1, ans2, ans3, ans4;` ` `  `    ``ans1 = ans2 = ans3 = ans4 = (``int``)1e9;` ` `  `    ``// vertically up` `    ``if` `(mat[n - ``1``][m] == ``0``) {` `        ``if` `(!vis[n - ``1``][m])` `            ``ans1 = ``1` `+ findMinSteps(mat, n - ``1``, m, dp, vis);` `    ``}` ` `  `    ``// horizontally right` `    ``if` `(mat[n][m + ``1``] == ``0``) {` `        ``if` `(!vis[n][m + ``1``])` `            ``ans2 = ``1` `+ findMinSteps(mat, n, m + ``1``, dp, vis);` `    ``}` ` `  `    ``// horizontally left` `    ``if` `(mat[n][m - ``1``] == ``0``) {` `        ``if` `(!vis[n][m - ``1``])` `            ``ans3 = ``1` `+ findMinSteps(mat, n, m - ``1``, dp, vis);` `    ``}` ` `  `    ``// vertically down` `    ``if` `(mat[n + ``1``][m] == ``0``) {` `        ``if` `(!vis[n + ``1``][m])` `            ``ans4 = ``1` `+ findMinSteps(mat, n + ``1``, m, dp, vis);` `    ``}` ` `  `    ``// minimum of every path` `    ``dp[n][m] = Math.min(ans1, Math.min(ans2, Math.min(ans3, ans4)));` `    ``return` `dp[n][m];` `}` ` `  `// Function that returns the minimum steps` `static` `int` `minimumSteps(``int` `mat[][], ``int` `n, ``int` `m)` `{` `    ``// index to store the location at` `    ``// which you are standing` `    ``int` `twox = -``1``;` `    ``int` `twoy = -``1``;` ` `  `    ``// find '2' in the matrix` `    ``for` `(``int` `i = ``0``; i < n; i++) {` `        ``for` `(``int` `j = ``0``; j < m; j++) {` `            ``if` `(mat[i][j] == ``2``) {` `                ``twox = i;` `                ``twoy = j;` `                ``break``;` `            ``}` `        ``}` `        ``if` `(twox != -``1``)` `            ``break``;` `    ``}` ` `  `    ``// Initialize dp matrix with -1` `    ``int` `dp[][]=``new` `int``[r][r];` `    ``for``(``int` `j=``0``;j= 1e9)` `        ``return` `-``1``;` `    ``else` `        ``return` `res;` `}` ` `  `// Driver Code` `public` `static` `void` `main(String args[])` `{` `    ``int` `mat[][] = { { ``1``, ``1``, ``1``, ``0``, ``1` `},` `                      ``{ ``1``, ``0``, ``2``, ``0``, ``1` `},` `                      ``{ ``0``, ``0``, ``1``, ``0``, ``1` `},` `                      ``{ ``1``, ``0``, ``1``, ``1``, ``0` `} };` ` `  `    ``System.out.println( minimumSteps(mat, r, c));` `}` `}` `//contributed by Arnab Kundu`

## Python

 `# Python program to find Minimum steps` `# to reach any of the boundary` `# edges of a matrix`   `r``=``4` `col``=``5` ` `  `# Function to find out minimum steps` `def` `findMinSteps(mat, n, m, dp,vis):`   `    ``# boundary edges reached` `    ``if` `(n ``=``=` `0` `or` `m ``=``=` `0` `or` `n ``=``=` `(r ``-` `1``) ``or` `m ``=``=` `(col ``-` `1``)):` `        ``return` `0` `    `  ` `  `    ``# already had a route through this` `    ``# point, hence no need to re-visit` `    ``if` `(dp[n][m] !``=` `-``1``):` `        ``return` `dp[n][m]` ` `  `    ``# visiting a position` `    ``vis[n][m] ``=` `True` ` `  `    ``ans1, ans2, ans3, ans4``=``10``*``*``9``,``10``*``*``9``,``10``*``*``9``,``10``*``*``9`   ` `  `    ``# vertically up` `    ``if` `(mat[n ``-` `1``][m] ``=``=` `0``):` `        ``if` `(vis[n ``-` `1``][m]``=``=``False``):` `            ``ans1 ``=` `1` `+` `findMinSteps(mat, n ``-` `1``, m, dp, vis)` `    `  ` `  `    ``# horizontally right` `    ``if` `(mat[n][m ``+` `1``] ``=``=` `0``):` `        ``if` `(vis[n][m ``+` `1``]``=``=``False``):` `            ``ans2 ``=` `1` `+` `findMinSteps(mat, n, m ``+` `1``, dp, vis)` `    `  ` `  `    ``# horizontally left` `    ``if` `(mat[n][m ``-` `1``] ``=``=` `0``):` `        ``if` `(vis[n][m ``-` `1``]``=``=``False``):` `            ``ans3 ``=` `1` `+` `findMinSteps(mat, n, m ``-` `1``, dp, vis)` `    `  ` `  `    ``# vertically down` `    ``if` `(mat[n ``+` `1``][m] ``=``=` `0``):` `        ``if` `(vis[n ``+` `1``][m]``=``=``False``):` `            ``ans4 ``=` `1` `+` `findMinSteps(mat, n ``+` `1``, m, dp, vis)` `    `  ` `  `    ``# minimum of every path` `    ``dp[n][m] ``=` `min``(ans1, ``min``(ans2, ``min``(ans3, ans4)))` `    ``return` `dp[n][m]`   ` `  `# Function that returns the minimum steps` `def` `minimumSteps(mat, n, m):`   `    ``# index to store the location at` `    ``# which you are standing` `    ``twox ``=` `-``1` `    ``twoy ``=` `-``1` ` `  `    ``# find '2' in the matrix` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(m): ` `            ``if` `(mat[i][j] ``=``=` `2``):` `                ``twox ``=` `i` `                ``twoy ``=` `j` `                ``break` `            `  `        `  `        ``if` `(twox !``=` `-``1``):` `            ``break` `    `  ` `  `    ``# Initialize dp matrix with -1` `    ``dp``=``[[``-``1` `for` `i ``in` `range``(col)] ``for` `i ``in` `range``(r)]` `    `  ` `  `    ``# Initialize vis matrix with false` `    ``vis``=``[[``False` `for` `i ``in` `range``(col)] ``for` `i ``in` `range``(r)]` `    `  ` `  `    ``# Call function to find out minimum steps` `    ``# using memoization and recursion` `    ``res ``=` `findMinSteps(mat, twox, twoy, dp, vis)` ` `  `    ``# if not possible` `    ``if` `(res >``=` `10``*``*``9``):` `        ``return` `-``1` `    ``else``:` `        ``return` `res`   ` `  `# Driver Code`     `mat``=` `[ [ ``1``, ``1``, ``1``, ``0``, ``1` `],` `      ``[ ``1``, ``0``, ``2``, ``0``, ``1` `],` `      ``[ ``0``, ``0``, ``1``, ``0``, ``1` `],` `      ``[ ``1``, ``0``, ``1``, ``1``, ``0` `] ]`   `print``(minimumSteps(mat, r, col))`   `#this is contributed by Mohit kumar 29`

## C#

 `// C# program to find Minimum steps ` `// to reach any of the boundary ` `// edges of a matrix`   `using` `System;`   `class` `Solution ` `{ ` `static` `int` `r=4,c=5; `   `    ``// Function to find out minimum steps ` `    ``static` `int` `findMinSteps(``int` `[,]mat, ``int` `n, ``int` `m, ``int` `[,]dp, ``bool` `[,]vis) ` `    ``{ ` `        ``// boundary edges reached ` `        ``if` `(n == 0 || m == 0 || n == (r - 1) || m == (c - 1)) { ` `            ``return` `0; ` `        ``} ` `    `  `        ``// already had a route through this ` `        ``// point, hence no need to re-visit ` `        ``if` `(dp[n,m] != -1) ` `            ``return` `dp[n,m]; ` `    `  `        ``// visiting a position ` `        ``vis[n,m] = ``true``; ` `    `  `        ``int` `ans1, ans2, ans3, ans4; ` `    `  `        ``ans1 = ans2 = ans3 = ans4 = (``int``)1e9; ` `    `  `        ``// vertically up ` `        ``if` `(mat[n - 1,m] == 0) { ` `            ``if` `(!vis[n - 1,m]) ` `                ``ans1 = 1 + findMinSteps(mat, n - 1, m, dp, vis); ` `        ``} ` `    `  `        ``// horizontally right ` `        ``if` `(mat[n,m + 1] == 0) { ` `            ``if` `(!vis[n,m + 1]) ` `                ``ans2 = 1 + findMinSteps(mat, n, m + 1, dp, vis); ` `        ``} ` `    `  `        ``// horizontally left ` `        ``if` `(mat[n,m - 1] == 0) { ` `            ``if` `(!vis[n,m - 1]) ` `                ``ans3 = 1 + findMinSteps(mat, n, m - 1, dp, vis); ` `        ``} ` `    `  `        ``// vertically down ` `        ``if` `(mat[n + 1,m] == 0) { ` `            ``if` `(!vis[n + 1,m]) ` `                ``ans4 = 1 + findMinSteps(mat, n + 1, m, dp, vis); ` `        ``} ` `    `  `        ``// minimum of every path ` `        ``dp[n,m] = Math.Min(ans1, Math.Min(ans2, Math.Min(ans3, ans4))); ` `        ``return` `dp[n,m]; ` `    ``} ` `    `  `    ``// Function that returns the minimum steps ` `    ``static` `int` `minimumSteps(``int` `[,]mat, ``int` `n, ``int` `m) ` `    ``{ ` `        ``// index to store the location at ` `        ``// which you are standing ` `        ``int` `twox = -1; ` `        ``int` `twoy = -1; ` `    `  `        ``// find '2' in the matrix ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``for` `(``int` `j = 0; j < m; j++) { ` `                ``if` `(mat[i,j] == 2) { ` `                    ``twox = i; ` `                    ``twoy = j; ` `                    ``break``; ` `                ``} ` `            ``} ` `            ``if` `(twox != -1) ` `                ``break``; ` `        ``} ` `    `  `        ``// Initialize dp matrix with -1 ` `        ``int` `[,]dp = ``new` `int``[r,r]; ` `        ``for``(``int` `j=0;j= 1e9) ` `            ``return` `-1; ` `        ``else` `            ``return` `res; ` `    ``} ` `    `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[,]mat = { { 1, 1, 1, 0, 1 }, ` `                        ``{ 1, 0, 2, 0, 1 }, ` `                        ``{ 0, 0, 1, 0, 1 }, ` `                        ``{ 1, 0, 1, 1, 0 }, }; ` `    `  `        ``Console.WriteLine(minimumSteps(mat, r, c)); ` `    ``} ` `    ``// This code is contributed by Ryuga`   `} `

## Javascript

 ``

Output:

`2`

Time Complexity: O(N^2)
Auxiliary Space: O(N^2)
Minimum steps to reach any of the boundary edges of a matrix | Set-2

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